Simplify; express your answer in exponential form. Assume $k\neq 0, y\neq 0$. $\dfrac{{(k^{-2})^{-5}}}{{(k^{5}y^{-3})^{2}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{-2}}$ to the exponent ${-5}$ . Now ${-2 \times -5 = 10}$ , so ${(k^{-2})^{-5} = k^{10}}$ In the denominator, we can use the distributive property of exponents. ${(k^{5}y^{-3})^{2} = (k^{5})^{2}(y^{-3})^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{-2})^{-5}}}{{(k^{5}y^{-3})^{2}}} = \dfrac{{k^{10}}}{{k^{10}y^{-6}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{10}}}{{k^{10}y^{-6}}} = \dfrac{{k^{10}}}{{k^{10}}} \cdot \dfrac{{1}}{{y^{-6}}} = k^{{10} - {10}} \cdot y^{- {(-6)}} = y^{6}$.